A lower bound for the equilateral number of normed spaces
Abstract
We show that if the Banach-Mazur distance between an n-dimensional normed space X and ℓ n∞ is at most 3/2, then there exist n + 1 equidistant points in X. By a well-known result of Alon and Milman, this implies that an arbitrary n-dimensional normed space admits at least e c √ log n equidistant points, where c > 0 is an absolute constant. We also show that there exist n equidistant points in spaces sufficiently close to ℓ n p , 1 < p < ∞.
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arXiv:math/0603614v1 [math.MG] 27 Mar 2006 A LOWER BOUND FOR THE EQUILATERAL NUMBER OF NORMED SPACES KONRAD J. SWANEPOEL AND RAFAEL VILLA Abstract. We show that if the Banach-Mazur distance between an n-dimensional normed space Xand ℓn ∞is at most 3/2, then there exist n+ 1 equidistant points in X. By a well-known result of Alon and Milman, this implies that an arbitrary n-dimensional normed space admits at least ec√log nequidistant points, where c > 0 is an absolute constant. We also show that there exist nequidistant points in spaces sufficiently close to ℓn p, 1 < p < ∞. 1. Notation Throughout the paper we use the same symbol cfor different absolute positive constants. Let Xdenote a normed space of finite dimension dim X=n. Let e(X) denote the largest size of an equilateral set in X. As usual, the space ℓn p, 1 ≤p < ∞, is defined as Rnwith the norm k(x1, x2,...,xn)kp= (Pi+1n|xi|p)1/p, and ℓn ∞is Rnwith the norm k(x1, x2,...,xn)k∞= maxi|xi|. The Banach-Mazur distance between two n-dimensional normed spaces is defined as d(X, Y ) = infkTk kT−1k, where the infimum is taken over all linear, invertible operators T:X→Y. We say that Xis a (1 + ε)-copy of Yif d(X, Y )≤1 + ε. 2. The main theorems It is conjectured [4, 9, 11, 12, 16] that e(X)≥n+ 1 for all n-dimensional normed spaces X. This is known for n≤3 [12] but open for n≥4. It is true for spaces sufficiently close to Euclidean: Theorem 1 (Brass [3] & Dekster [6]).Let Xbe an n-dimensional normed space with Banach-Mazur distance d(X, ℓn 2)≤1 + 1 n. Then an equilateral set in Xof at most npoints can be extended to one of n+ 1 points. In particular, e(X)≥n+ 1. 2000 Mathematics Subject Classification. Primary 46B04; Secondary 46B20, 52A21, 52C17. This material is based upon work supported by the South African National Research Foundation under Grant number 2053752. The second author thanks the DGES grant BFM2003-01297 for financial support. Parts of this paper was written during a visit of the second author to the Department of Mathematical Sciences, University of South Africa, in January 2006. 1
2 KONRAD J. SWANEPOEL AND RAFAEL VILLA Combining this theorem with Gordon’s estimate [8] in the Dvoretzky theorem [7, §4], the following general lower bound follows: e(X)≥c(log n)1/3. We improve this to the following: Theorem A. For any n-dimensional normed space Xwe have e(X)≥ ec√log n, where c > 0is an absolute constant. The proof works by looking for large subspaces close to either ℓk 2or ℓk ∞. To this end we use the following theorem: Theorem 2 (Alon-Milman [1]).Let Xbe an n-dimensional normed space. Then for each ε > 0there exists c=c(ε)>0such that Xeither contains a (1 + ε)-isomorphic copy of ℓm 2for some msatisfying log log m≥1 2log log n or contains a (1+ε)-isomorphic copy of ℓk ∞for some ksatisfying log log k > 1 2log log n−c. If the second case occurs in the above theorem, i.e., if we find a (1 + ε)- isomorphic copy of ℓk ∞in X, then we need a result similar to Theorem 1 for spaces near ℓn ∞. This we provide as follows: Theorem B. Let Xbe an n-dimensional normed space with Banach-Mazur distance d(X, ℓn ∞)≤3/2. Then e(X)≥n+ 1. We can then choose ε= 1/2 in Theorem 2 to obtain e(X)≥ec√log nin this case. On the other hand, if we find a (1 + 1 2)-isomorphic copy of ℓm 2where m > e√log n, we cannot yet apply Theorem 1. We first have to find a kdimensional subspace of the m-dimensional space that is (1 + 1 k)-isomorphic to ℓk 2. This is provided by the following: Theorem 3 (Milman [10]).Let Xbe an m-dimensional normed space and 0< ε < 1. Then Xcontains a k-dimensional subspace Ywhere k≥cε2m/d2(X, ℓm 2)and d(Y, ℓk 2)≤1 + ε. See also [7, Corollary 4.2.2]. Putting ε= 1/k and Xthe m-dimensional subspace that is 3/2-isomorphic to ℓm 2into the above theorem, we obtain k > cm1/3, and Theorem 1 then gives e(X)> cm1/3> ce(1/3)√log n. To complete the proof of Theorem A, it only remains to prove Theorem B. Proof of Theorem B. We use the Brouwer fixed point theorem [5, §14.3], as in Brass’ proof of Theorem 1. Without loss of generality we may assume X= (Rn,k·k) and kxk ≤ kxk∞≤3 2kxkfor all x∈X.
A LOWER BOUND FOR THE EQUILATERAL NUMBER OF NORMED SPACES 3 Let I={(i, j) : 1 ≤i < j ≤n+ 1}, with |I|=n(n+ 1)/2 = N. For ε= (εi,j)(i,j)∈I∈[0,1 2]N, let p1(ε) = (−1,0,...0), pj(ε) = (ε1,j,...εj−1,j,−1,0,...0),2≤j≤n−1, pn(ε) = (ε1,n,...εn−1,n,−1), pn+1(ε) = (ε1,n+1,...εn,n+1). For 1 ≤i < j ≤nwe have kpi(ε)−pj(ε)k∞= 1 + εi,j. Define ϕ: [0,1/2]N→[0,1/2]Nby ϕi,j(ε) = 1 + εi,j − kpi(ε)−pj(ε)k,1≤i < j ≤n. Note that ϕi,j(ε)≥1 + εi,j − kpi−pjk∞= 0 and ϕi,j(ε)≤1 + εi,j −2 3kpi−pjk∞=1 3(1 + εi,j)≤1 2, so ϕis well-defined. Brouwer now gives the existence of a point ε′= (ε′ i,j)∈ [0,1/2]Nwith ϕ(ε′) = ε′, which implies that kpi(ε′)−pj(ε′)k= 1 for all 1≤i < j ≤m. We have obtained n+ 1 equilateral points. 3. A generalization to ℓn p The following theorem partially generalizes Theorem B to all ℓn pspaces with 1 < p < ∞. Theorem C. For each n > 2and p∈(1,∞)there exists R(p, n)>1such that for any n-dimensional normed space Xwith Banach-Mazur distance d(X, ℓn p)≤R(p, n)we have e(X)≥n. In fact, R(p, n) = max θ>01 + (1 + θ)p 2 + (n−2)θp1/p ∼1 + p−1 2pn−1 p−1as n→ ∞ with pfixed. Proof. We follow the proof of Theorem B. Assume X= (Rn,k·k) and kxk ≤ kxkp≤Rkxkfor all x∈X. Fix β, γ > 0. Let I={(i, j) : 1 ≤i < j ≤n}, with |I|=n(n−1)/2 = N. For ε= (εi,j)(i,j)∈I∈[0, β]N, let p1(ε) = (−γ, 0,...0), pj(ε) = (ε1,j,...εj−1,j,−γ, 0,...0),2≤j≤n−1, pn(ε) = (ε1,n,...εn−1,n,−γ). For 1 ≤i < j ≤nwe have kpj−pikp p= i−1 X k=1 |εk,j −εk,i|p+ (εi,j +γ)p+ j X k=i+1 εp k,j +γp.
4 KONRAD J. SWANEPOEL AND RAFAEL VILLA Define ϕ: [0, β]N→[0, β]Nby ϕi,j(ε) = 1+εi,j −kpi−pjkfor 1 ≤i < j ≤n. On the one hand, ϕi,j(ε)≤1 + εi,j −R−1kpi−pjkp ≤1 + εi,j −R−1[(γ+εi,j)p+γp]1/q . Taking into account that the latter is increasing with respect to εi,j, the inequality εi,j ≤βimplies ϕi,j(ε)≤1 + β−R−1[(γ+β)p+γp]1/q . Therefore, if (γ+β)p+γp≥Rpthen ϕi,j(ε)≤β. On the other hand ϕi,j(ε)≥1 + εi,j − kpi−pjkp ≥1 + εi,j −[(n−2)βp+ (γ+εi,j)p+γp]1/q . Again the latter is increasing with respect to εi,j, so using εi,j ≥0 we have ϕi,j(ε)≥1−[(n−2)βp+ 2γp]1/q . Then ϕi,j(ε1,...εm)≥0 would follow if (n−2)βp+ 2γp≤1.Subsequently, if (∗) (γ+β)p+γp≥Rpand (n−2)βp+ 2γp≤1, then ϕis well defined. Brouwer now gives a point ε′= (ε′ i,j)∈[0, β]Nsuch that ϕ(ε′) = ε′, implying that the points p1(ε′),...pn(ε′) are equilateral. Finally, to take the best choice for the parameters in (∗) we have to maximize the expression (γ+β)p+γpunder the constraints (n−2)βp+2γp≤ 1 and β, γ ≥0. Setting θ=β/γ, we obtain Rp= max θ>0 1 + (1 + θ)p 2 + (n−2)θp. It is not difficult to see that for θclose to n−1/(p−1) the right-hand side is >1 and R−1∼p−1 2pn−1 p−1. 4. Concluding remarks For p= 2 the estimate in the above theorem is d(X, ℓn 2).1+ 1 4n, slightly worse than Theorem 1. However, we don’t know how to obtain n+1 equidistant points as in Theorem B. It would also be interesting to know whether arbitrary equilateral sets of at most npoints in spaces near ℓn pcan be extended as in Theorem 1. A different idea will be needed to extend the above theorem to the case p= 1. See [15] for a survey on equilateral sets, as well as [2, 13, 14] for further results on equilateral sets in ℓn p.
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